AP Calculus BC

The Ratio Test: How It Works and When to Use It

Q: What does it mean when the ratio test equals 1?

It means the test is inconclusive: it gives you no information, and the series could still converge or diverge. Do not read L = 1 as divergence. Switch to another test. For rational terms or a p-series use a p-series argument or the limit comparison test; for an alternating leftover use the alternating series test; for the endpoints of a power series test each endpoint on its own.

Q: Does the ratio test work on alternating series?

Yes. The test uses the absolute value \left| \frac{a_{n+1}}{a_n} \right|, so signs do not matter while you compute L. If L < 1, the series converges absolutely, which means it converges regardless of the alternating signs (Topic 10.9). If L = 1, fall back to the alternating series test.

Q: When should I use the ratio test instead of the comparison test?

Use the ratio test when the term has factorials or something raised to the n-th power, since those cancel cleanly in the ratio. Use a comparison or limit comparison test when the term is a rational function of n (polynomials over polynomials), because there the ratio test just gives L = 1 and tells you nothing.

Q: Is the root test on the AP Calculus BC exam?

No. The CED lists the tests assessed on the BC exam (nth term, integral, comparison, limit comparison, alternating series, and ratio), and the root test is not among them. It exists and can handle some series the ratio test cannot, but it is beyond the AP scope, so stick with the ratio test for exam work.

Compute L, the limit as n approaches infinity of the absolute value of a term over the one before it. If L < 1 the series converges absolutely; if L > 1 or infinite it diverges; if L = 1 it is inconclusive, so switch tests. Use the ratio test when terms have factorials or a base to the n-th power like r^n.

The ratio test in one move

The ratio test looks at how fast a series' terms shrink by comparing each term to the one right before it. Take the absolute value of the ratio of consecutive terms and send $n$ to infinity to get a single number $L$ .

L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|

That one number decides everything:

$L < 1$ : the series converges absolutely.
$L > 1$ (including $L = \infty$ ): the series diverges.
$L = 1$ : the test is inconclusive and tells you nothing, so you must use a different test.

This is CED Topic 10.8, BC only. Notice the absolute value: the ratio test always checks for absolute convergence, so it works directly on series with negative or alternating terms (Topic 10.9). If $L < 1$ , the series converges no matter what the signs are doing.

Knowing when to reach for it

Method selection matters more than mechanics here. The ratio test is built to cancel two structures: factorials and expressions raised to the $n$ -th power. When you see an $n!$ anywhere in the term, the ratio test is almost always the fastest route, because $\frac{(n+1)!}{n!} = n+1$ collapses the factorial to a single factor. The same thing happens with an exponential like $r^n$ , since $\frac{r^{n+1}}{r^n} = r$ .

So the recognition trigger is simple: scan the general term $a_n$ . If it contains a factorial, a constant base raised to the $n$ (like $2^n$ or $5^n$ ), or a product of both, start with the ratio test. A term like $n^n$ also triggers the ratio test, but note its ratio does not collapse to a constant, since $\frac{(n+1)^{n+1}}{n^n} \to \infty$ .

Just as useful is knowing when it will fail. If the term is purely algebraic, a ratio of polynomials in $n$ such as $\frac{1}{n^2}$ or $\frac{n}{n^3+1}$ , the ratio of consecutive terms drifts to $1$ and the test is inconclusive. A p-series behaves the same way. For those, skip the ratio test and go straight to a p-series argument, the comparison or limit comparison test, or the integral test.

Recognition rule

Factorials or something to the $n$ -th power: use the ratio test. Rational functions of $n$ or p-series: expect $L = 1$ and use a different test.

Running the test step by step

Once you have decided the ratio test fits, the procedure is mechanical.

Write the general term $a_n$ clearly, then write $a_{n+1}$ by replacing every $n$ with $n+1$ .
Form the ratio $\frac{a_{n+1}}{a_n}$ and take its absolute value.
Simplify by multiplying by the reciprocal of $a_n$ : flip the denominator and cancel. Use $(n+1)! = (n+1)\cdot n!$ to reduce factorials and $\frac{r^{n+1}}{r^n} = r$ to reduce powers.
Take the limit as $n \to \infty$ to find $L$ .
Compare $L$ to $1$ and state the conclusion.

The single most useful identity is $(n+1)! = (n+1)\cdot n!$ . It lets the $n!$ cancel with the $(n+1)!$ and leaves just the factor $n+1$ , which is exactly why factorial series come out so clean under this test.

What to do when L = 1

An inconclusive result is not a divergence result. $L = 1$ means the ratio test cannot see the behavior of the series, not that the series fails to converge. You have to reach for another tool.

Two situations produce $L = 1$ in practice. The first is algebraic terms and p-series, where the right follow-up is a p-series argument or the limit comparison test. If the leftover series alternates, the alternating series test usually finishes the job; if it is positive with polynomial terms, comparison or the integral test does.

The second is the endpoints of a power series. The ratio test hands you the radius of convergence, but at each endpoint the ratio limit lands on $1$ , so you test those two $x$ -values separately with whatever ordinary series test fits (Topic 10.13). The worked example on the radius and interval of convergence below shows this in full.

Common mistakes

Dropping the absolute value. The test is stated with $\left| \cdot \right|$ ; without it a negative ratio can produce a meaningless limit.
Forgetting to flip. Dividing by $a_n$ means multiplying by its reciprocal, so a common slip is to multiply by $a_n$ instead.
Mishandling factorials. $\frac{(n+1)!}{n!} = n+1$ , not $1$ and not another factorial.
Calling $L = 1$ a divergence. It is inconclusive, so switch tests.
Skipping endpoints. After the ratio test gives a radius, a power series is not finished until both endpoints are checked separately.

Worked examples

Worked example

Factorial in the denominator

Determine whether $\sum_{n=1}^{\infty} \frac{2^n}{n!}$ converges or diverges.

The term has a factorial, so the ratio test is the natural choice. Set $a_n = \frac{2^n}{n!}$ and $a_{n+1} = \frac{2^{n+1}}{(n+1)!}$ .
Form the absolute ratio and flip the denominator: $\left| \frac{a_{n+1}}{a_n} \right| = \frac{2^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n}$ .
Cancel using $\frac{2^{n+1}}{2^n} = 2$ and $\frac{n!}{(n+1)!} = \frac{1}{n+1}$ : the ratio becomes $2 \cdot \frac{1}{n+1} = \frac{2}{n+1}$ .
Take the limit: $\lim_{n \to \infty} \frac{2}{n+1} = 0$ , so $L = 0$ .
Since $L = 0 < 1$ , the series converges absolutely.

The series converges absolutely, since $L = 0 < 1$ .

Worked example

Factorial beats a big exponential

Determine whether $\sum_{n=1}^{\infty} \frac{n!}{100^n}$ converges or diverges.

A factorial signals the ratio test. Let $a_n = \frac{n!}{100^n}$ , so $a_{n+1} = \frac{(n+1)!}{100^{n+1}}$ .
Form the absolute ratio: $\left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)!}{100^{n+1}} \cdot \frac{100^n}{n!}$ .
Cancel with $\frac{(n+1)!}{n!} = n+1$ and $\frac{100^n}{100^{n+1}} = \frac{1}{100}$ : the ratio is $\frac{n+1}{100}$ .
Take the limit: $\lim_{n \to \infty} \frac{n+1}{100} = \infty$ , so $L = \infty$ .
Since $L > 1$ , the series diverges. The takeaway: a factorial in the numerator overwhelms even a base as large as $100$ , because the ratio grows without bound.

The series diverges, since $L = \infty > 1$ .

Worked example

Radius and interval of convergence

Find all $x$ for which the power series $\sum_{n=1}^{\infty} \frac{x^n}{n}$ converges.

This is Topic 10.13: apply the ratio test to the terms, treating $x$ as fixed. Let $a_n = \frac{x^n}{n}$ , so $a_{n+1} = \frac{x^{n+1}}{n+1}$ .
Form the absolute ratio: $\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x^{n+1}}{n+1} \cdot \frac{n}{x^n} \right| = |x| \cdot \frac{n}{n+1}$ .
Take the limit: $\lim_{n \to \infty} |x| \cdot \frac{n}{n+1} = |x|$ , so $L = |x|$ .
The ratio test guarantees convergence when $L < 1$ , that is $|x| < 1$ . The radius of convergence is $R = 1$ , giving the open interval $-1 < x < 1$ .
At the endpoints $L = 1$ , so the ratio test is inconclusive and each endpoint is tested on its own. At $x = 1$ the series is $\sum \frac{1}{n}$ , the harmonic series, which diverges. At $x = -1$ the series is $\sum \frac{(-1)^n}{n}$ , the alternating harmonic series, which converges by the alternating series test.
Combine the results: the series converges for $-1 \le x < 1$ .

Converges on $[-1, 1)$ : radius $R = 1$ , diverging at $x = 1$ and converging at $x = -1$ .

Frequently asked questions

What does it mean when the ratio test equals 1?

It means the test is inconclusive: it gives you no information, and the series could still converge or diverge. Do not read $L = 1$ as divergence. Switch to another test. For rational terms or a p-series use a p-series argument or the limit comparison test; for an alternating leftover use the alternating series test; for the endpoints of a power series test each endpoint on its own.

Does the ratio test work on alternating series?

Yes. The test uses the absolute value $\left| \frac{a_{n+1}}{a_n} \right|$ , so signs do not matter while you compute $L$ . If $L < 1$ , the series converges absolutely, which means it converges regardless of the alternating signs (Topic 10.9). If $L = 1$ , fall back to the alternating series test.

When should I use the ratio test instead of the comparison test?

Use the ratio test when the term has factorials or something raised to the $n$ -th power, since those cancel cleanly in the ratio. Use a comparison or limit comparison test when the term is a rational function of $n$ (polynomials over polynomials), because there the ratio test just gives $L = 1$ and tells you nothing.

Is the root test on the AP Calculus BC exam?

No. The CED lists the tests assessed on the BC exam (nth term, integral, comparison, limit comparison, alternating series, and ratio), and the root test is not among them. It exists and can handle some series the ratio test cannot, but it is beyond the AP scope, so stick with the ratio test for exam work.